Proof of all submodules of $M$ are f.g. then $M$ is Noetherian I've been thinking about the following proof that if all submodules of $M$ are finitely generated then $M$ is Noetherian:
Assume $M$ is such that all submodules are finitely generated. Then in particular, $M$ is finitely generated. Now let $N_1 \subset N_2 \subset \dots $ be an increasing chain of submodules in $M$. If the chain is non-stabilising, then for $k$ large enough $N_k$ will eventually contain all generators of $M$. Hence $N_k = N_{k+1} = \dots $.
I think the idea is correct but on the other hand, it sort of doesn't feel right. Is this proof correct? Or if not: can it be fixed? Thank you.
 A: The standard three properties are:

Theorem. Let $R$ be a ring, and let $M$ be an $R$-module. The following are equivalent:
  
  
*
  
*$M$ has ACC on submodules; that is, if
  $$N_1\subseteq N_2\subseteq\cdots\subseteq N_m\subseteq\cdots$$
  is an ascending chain of submodules of $M$, then the chain stabilizes; that is, there exists $m_0$ such that $N_{m_0}=N_{m_0+k}$ for all $k\geq 0$.
  
*Every submodule of $M$ is finitely generated.
  
*Every nonempty collection of submodules of $M$ has maximal elements (under inclusion).

Some of the implications require the Axiom of Choice. In the absence of AC, the strongest condition is 3, which implies the other 2 without having to invoke AC.
Proof. 1$\implies$2 (This implication requires at least some Choice). We prove it by contrapositive. Assume there is submodule $N$ of $M$ that is not finitely generated. Then $N\neq 0$, so let $n_1\in N$, $n_1\neq 0$. Let $N_1=\langle n_1\rangle$; then $N_1\neq N$, since $N$ is not finitely generated, so there exists $n_2\in N\setminus N_1$. Let $N_2 = \langle n_1,n_2\rangle$. Then $N_1\subsetneq N_2$.
Assume we have chosen $n_1,\ldots,n_k$, with $N_i = \langle n_1,\ldots,n_i\rangle$, and with $N_i\subsetneq N_{i+1}$, $i=1,\ldots,k-1$. Since $N$ is not finitely generated, $N\setminus N_{k}$ is nonempty; let $n_{k+1}\in N\setminus N_{k}$. Then $N_k\subsetneq \langle n_1,\ldots,n_{k},n_{k+1}\rangle = N_{k+1}$.
Inductively (this requires at least dependent countable choice), we obtain an infinite ascending chain that does not stabilizer,
$$N_1\subsetneq N_2\subsetneq \cdots\subsetneq N_m\subsetneq N_{m+1}\subsetneq \cdots.$$
2$\implies$1 This implication holds without the Axiom of Choice, and is the one you are working on:
Let $N_1\subseteq N_2\subseteq\cdots\subseteq N_m\subseteq$ be an ascending chain of submodules. It is easy to verify that 
$$N = \bigcup_{i=1}^{\infty}N_i$$
is a submodule of $M$. By assumption, $N$ is finitely generated; let $n_1,\ldots,n_k$ be a generating set. For each $i$, there exists $m_i$ such that $n_i\in N_{m_i}$. Let $m=\max\{m_1,\ldots,m_k\}$. Then $N_{m_i}\subseteq N_m$, hence $n_i\in N_{m_i}\subseteq N_m$. Thus, $N = \langle n_1,\ldots,n_k\rangle\subseteq N_m\subseteq N_{m+r}\subseteq N$ for all $r\geq 0$, hence $N_m=N_{m+k}=N$ for all $k\geq 0$. Thus, the chain stabilizes.
2$\implies$3. This follows from Zorn's Lemma. Consider a nonempty collection $$\mathcal{S}$ of submodules of $M$, ordered by inclusion. We prove that it satisfies the hypothesis of Zorn's Lemma, and hence has maximal elements. Let $\mathcal{C}$ be a chain of elements of $\mathcal{S}$; then $\cup\mathcal{C}$ is a submodule of $M$, and hence is finitely generated; therefore, there exist $m_1,\ldots,m_n$ that generate $\cup\mathcal{C}$; hence there exist $M_1,\ldots,M_n\in \mathcal{C}$ such that $m_i\in M_i$. Since $\mathcal{C}$ is a chain, one of $M_1,\ldots,M_n$ contains all the others (finite subsets of a chain always have a maximum); say it is $M_0$. Then $\cup \mathcal{C}=M$. Hence, if $N$ is an element of $\mathcal{C}$, then $N\subseteq\cup\mathcal{C}=M_0\in\mathcal{S}$. Thus, $\mathcal{C}$ has an upper bound in $\mathcal{S}$. 
Since every chain in $\mathcal{S}$ is bounded above in $\mathcal{S}$, by Zorn's Lemma $\mathcal{S}$ has maximal elements, as desired.
3$\implies$1 This implication holds without AC. Given an ascending chain of submodules of $M$, let $\mathcal{S}$ be the collection of these modules. By hypothesis, $\mathcal{S}$ has a maximal element, $N_m$. Since $\mathcal{S}$ is a chain, a maximal element must be a maximum, so $N_m$ contains all other elements of the chain. In particular, if $k\geq 0$, then $N_{m+k}\subseteq N_m\subseteq N_{m+k}$, hence $N_m=N_{m+k}$, as desired.
3$\implies$2 This implication holds without AC. Let $N$ be a submodule of $M$. Let $\mathcal{S}$ be the collection of all finitely generated submodules of $N$, ordered by inclusion. Since $0\in\mathcal{S}$, the latter is nonempty. By hypothesis, $\mathcal{S}$ has maximal elements. Let $N_0$ be a maximal element. Now, $N_0\subseteq N$ by assumption. Let $x\in N$. Then $\langle N_0,x\rangle$ is a finitely generated submodule of $N$, and clearly $N_0\subseteq \langle N_0,x\rangle$. By the maximality of $N_0$, we conclude that $N_0=\langle N_0,x\rangle$, hence $x\in N_0$; thus, $N\subseteq N_0$, proving equality, and hence that $N$ is finitely generated, as desired. $\Box$
I don't know a direct proof of 1$\implies 3$ that does not go through 2, but in any case the implication requires the Axiom of Choice, as far as I know.
A: I think you can formalise this a bit more. For example, you can start of by saying suppose we have an increasing chain of submodules $N_1 \subseteq N_2 \subseteq N_3 \subseteq \ldots $ Then you can check that 
$$N = \bigcup_{i \geq 1} N_i$$
is a submodule of $M$. Now by assumption all submodules of $M$ are finitely generated so that $N$ is finitely generated. Now how does this tell me that the chain is eventually constant?
